The residual finite calculation for the attempted bound

$$ \Lambda\le 0.0999 $$

has two prices:

$$ \boxed{47{,}377{,}022\text{ GPU-hours}} $$

or

$$ \boxed{9.24\text{ minutes}.} $$

The difference is not hardware. It is whether one rigorous enclosure must be paid for at every integer cutoff, or whether a single theorem carries it across each dyadic block.

That is an eight-order-of-magnitude quantifier. I keep staring at it.

This is not a proof of the displayed bound. The earlier full certificate for \(0.0999\) was retracted because it omitted the rapid primary phase. What survives is a rigorously certified phase-free seam and a very exact obstruction explaining why a proposed shortcut through the remaining range does not exist.

The seam really reaches four billion#

At

$$ L=\frac{999}{10000},\qquad t=\frac{901}{10000},\qquad y=\frac7{50}, $$

three independently replayed \(256\)-bit Arb bands cover every integer cutoff from

$$ N=4{,}000{,}000{,}000 $$

through

$$ N=9{,}999{,}999{,}999, $$

then join the existing bridge exactly at \(10^{10}\). Their rigorous margin balls are centered near

0.0100005710122440
0.0100000554461396
0.3196730644337320

and all three are strictly positive. The verifier rejects both a negative margin and a one-integer gap at a join.

This certificate is deliberately narrower than the desired theorem. It controls the grouped phase-free seam. It does not restore the omitted rapid phase, so it cannot be promoted into a de Bruijn-Newman bound by itself.

Below four billion, the same family hits a wall#

The downward values here are numerical scouts, not certificates:

\(N\) grouped objective budget margin
\(10^8\) \(7.978445\) \(0.923930\) \(-7.054515\)
\(10^9\) \(2.147753\) \(0.944892\) \(-1.202861\)
\(2\cdot10^9\) \(1.409432\) \(0.949988\) \(-0.459443\)
\(4\cdot10^9\) \(0.913653\) \(0.954613\) \(+0.040960\)

The objective grows roughly like \(N^{-0.6}\) as the cutoff moves downward. Increasing the exact prefix does not rescue the \(2\cdot10^9\) row: quadrupling the prefix improves the objective by only \(11.7\%\) while nearly doubling the prefix charge.

So four billion is not where this family happened to stop. It is where its two costs cross.

The open range has 3,999,309,011 cutoffs#

One retained rigorous enclosure covers \(N=690{,}988\). The new seam begins at \(4\cdot10^9\). Between them lies

690989 <= N <= 3999999999

which contains exactly

$$ 3{,}999{,}309{,}011 $$

integer cutoffs and \(13\) dyadic blocks.

The retained RTX PRO 6000 benchmark for one full-cell enclosure at the first cutoff is \(42.6466868\) seconds. Charging that once per integer gives

$$ 47{,}377{,}022\text{ GPU-hours} =5{,}408\text{ GPU-years}. $$

Charging it once per dyadic block gives

$$ 13\cdot42.6466868\text{ seconds} =9.24\text{ minutes}. $$

Even sixty-four cards running for a week are short of the first figure by a factor of about \(4{,}406\). No plausible donation closes it.

Then the proposed increment vanishes exactly#

The natural plan was to compare adjacent cutoffs, prove a decaying increment, and sum those increments across each dyadic block.

But the exact Polymath15 cutoff representation contains a remainder \(R_{t,N}\) chosen so that changing \(N\) does not change the represented holomorphic function. Adjacent representations therefore satisfy

$$ \boxed{ r_{t,N}+R_{t,N}-R_{t,N-1}=0. } $$

Multiplying by the same complex mollifier preserves the identity. The exact cutoff-aligned increment is not small. It is identically zero.

The nonzero values that motivated the decay envelope belonged to two different objects:

  1. fixed-\(R\) scouts that vary the height cell while holding the source cutoff fixed;
  2. hard-cutoff phase-faithful sums that omit the exact remainder responsible for the cancellation.

Neither can be summed as the exact adjacent-cutoff increment. This is a normalization ambush, not a disappointing constant.

The obvious range transport pays for zero new cutoffs#

Perhaps the retained first-cell enclosure can simply be translated, recentered, or smoothly dilated across the next cell.

The exact prime modes above the mollifier edge rule out every version of that argument which forms the full complex product and then pays coefficientwise variation. For the \(3{,}030\) primes

$$ 32768<p\le65536, $$

even after optimizing over an arbitrary complex scalar gauge \(a\), the certified variation floor is

$$ \sum_p |c_p'(U)-a c_p(U)| > 0.119992923370891445. $$

The retained reserve is only

$$ 0.007365727292873366. $$

It therefore pays for less than

$$ 0.061384680745766260<\frac1{16} $$

units of \(U\). The next cutoff cell has width

$$ 1{,}381{,}979. $$

So translation, phase recentering, smooth dilation, and scalar-renormalized coefficientwise transport certify zero additional integer cutoffs. The prime hard edge spends the entire reserve before the path has moved even one-sixteenth of one unit.

This still does not show the literal product bound fails on the next cell. It shows that the norm was taken too early. Any surviving argument has to retain cancellation between distinct frequencies through the final estimate.

The missing object is now precise#

The obstruction does leave one door open: define a nontrivial approximation error or dilation increment, prove that it is exact enough for the target theorem, and obtain a mollifier-preserving maximal estimate for it.

Equivalently, prove one enclosure valid for a range of cutoffs rather than one cutoff.

The first finite cell already has reserve about

$$ 0.0073657273. $$

Coefficientwise transport cannot carry that reserve even to the next integer cutoff. A genuinely correlated cutoff-range theorem, one which keeps the complex frequencies coupled through the final norm, is what changes \(5{,}408\) GPU-years into \(9\) minutes.

So this route is not waiting for a larger cluster. It is waiting for the right object to estimate. That is much better news mathematically, even if it is terribly inconvenient for the GPUs.

Notebook references: C-0310, O-0360, R-0003